如何从嵌套在两个列表中的字典中获取值?

Question

我仍然熟悉python和使用API。我从 API 中提取了以下数据：

[[{'term': 'Boys', 'AAT_URL': 'http://vocab.getty.edu/page/aat/300247598'},
{'term': 'Fish', 'AAT_URL': 'http://vocab.getty.edu/page/aat/300266085'},
{'term': 'Boats', 'AAT_URL': 'http://vocab.getty.edu/page/aat/300178749'}],
[{'term': 'Bears', 'AAT_URL': 'http://vocab.getty.edu/page/aat/300266516'}],
None,
[{'term': 'Interiors',
'AAT_URL': 'http://vocab.getty.edu/page/aat/300391239'},
{'term': 'Jewelry', 'AAT_URL': 'http://vocab.getty.edu/page/aat/300209286'},
{'term': 'Couples', 'AAT_URL': 'http://vocab.getty.edu/page/aat/300379217'},
{'term': 'Men', 'AAT_URL': 'http://vocab.getty.edu/page/aat/300025928'},
{'term': 'Women', 'AAT_URL': 'http://vocab.getty.edu/page/aat/300025943'},
{'term': 'Weights and Measures',
'AAT_URL': 'http://vocab.getty.edu/page/aat/300386648'}],
...]

我想在最终列表中提取"term"键(但保留子列表(中的值，如下所示：

[['Boys','Fish','Boats'],['Bears'],['Interiors','Jewlery','Couples','Men','Women','Weights and Measures']...]

每当我尝试像这样浏览列表时：

for x in tags:
for y in x:
print (y['term'])

我得到以下结果：

Boys
Fish
Boats
Bears
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-48-f2ff632ad612> in <module>
11 
12 for x in tags:
---> 13     for y in x:
14         print (y['term'])
TypeError: 'NoneType' object is not iterable

我该如何解决？提前谢谢你 c：

Answer 1

在迭代之前，对照None检查x：

for x in tags:
if x is not None:
for y in x:
print (y['term'])

Answer 2

如果您看到从 API 中提取的数据中还有一个元素 None。因此，您必须检查在迭代时不会遇到这种情况。

for x in tags:
if(x is not None):
for y in x:
print(y['term'])

这将解决问题

Answer 3

要将数据提取到相同结构的列表中，请检查None，并且仅在有数据时才添加内容：

data = [[{'term': 'Boys', 'AAT_URL': 'http://vocab.getty.edu/page/aat/300247598'},
{'term': 'Fish', 'AAT_URL': 'http://vocab.getty.edu/page/aat/300266085'},
{'term': 'Boats', 'AAT_URL': 'http://vocab.getty.edu/page/aat/300178749'}],
[{'term': 'Bears', 'AAT_URL': 'http://vocab.getty.edu/page/aat/300266516'}], 
None,
[{'term': 'Interiors', 'AAT_URL': 'http://vocab.getty.edu/page/aat/300391239'},
{'term': 'Jewelry', 'AAT_URL': 'http://vocab.getty.edu/page/aat/300209286'},
{'term': 'Couples', 'AAT_URL': 'http://vocab.getty.edu/page/aat/300379217'},
{'term': 'Men', 'AAT_URL': 'http://vocab.getty.edu/page/aat/300025928'},
{'term': 'Women', 'AAT_URL': 'http://vocab.getty.edu/page/aat/300025943'},
{'term': 'Weights and Measures', 'AAT_URL': 'http://vocab.getty.edu/page/aat/300386648'}]]
result = []
for inner_list in data:
if inner_list:  # None and [] are falsy, ignore them
# got something, add a new empty list to add things into
result.append([])
for d in inner_list:
# add all dicts term to the last item of result
result[-1].append(d["term"])  
print(result)

将输出：

[['Boys', 'Fish', 'Boats'], ['Bears'], 
['Interiors', 'Jewelry', 'Couples', 'Men', 'Women', 'Weights and Measures']]

Answer 4

首先过滤掉 None ，然后使用映射函数

data = [
[
{"term": "Boys", "AAT_URL": "http://vocab.getty.edu/page/aat/300247598"},
{"term": "Fish", "AAT_URL": "http://vocab.getty.edu/page/aat/300266085"},
{"term": "Boats", "AAT_URL": "http://vocab.getty.edu/page/aat/300178749"},
],
[{"term": "Bears", "AAT_URL": "http://vocab.getty.edu/page/aat/300266516"}],
None,
[
{"term": "Interiors", "AAT_URL": "http://vocab.getty.edu/page/aat/300391239"},
{"term": "Jewelry", "AAT_URL": "http://vocab.getty.edu/page/aat/300209286"},
{"term": "Couples", "AAT_URL": "http://vocab.getty.edu/page/aat/300379217"},
{"term": "Men", "AAT_URL": "http://vocab.getty.edu/page/aat/300025928"},
{"term": "Women", "AAT_URL": "http://vocab.getty.edu/page/aat/300025943"},
{"term": "Weights and Measures", "AAT_URL": "http://vocab.getty.edu/page/aat/300386648"},
],
]

result = map(lambda x: list(map(lambda i: i["term"], x)), filter(lambda x: x is not None, data))
print(list(result))